Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 100005
using namespace std;
typedef long long ll;
int n,b[N],m,t[N<<3];
ll K,a[N],s[N],L[N],R[N],v[N<<1],f[N];
void update(int k,int l,int r,int L,int R,int val){
	if(L>R) return;
	if(L<=l&&R>=r){
		t[k]=val;
		return;
	}
	if(t[k]) t[k<<1]=t[k<<1|1]=t[k],t[k]=0;
	int mid=l+r>>1;
	if(L<=mid) update(k<<1,l,mid,L,R,val);
	if(R>mid) update(k<<1|1,mid+1,r,L,R,val);
}
int query(int k,int l,int r,int x){
	if(l==r) return t[k];
	if(t[k]) t[k<<1]=t[k<<1|1]=t[k],t[k]=0;
	int mid=l+r>>1;
	if(x<=mid) return query(k<<1,l,mid,x);
	else return query(k<<1|1,mid+1,r,x);
}
ll calc(int x){
	int y=query(1,1,m,x);
	if(!y) return 0;
	return f[y]+(v[L[y]]-v[x]+K)%K;
}
int main(){
	scanf("%d%lld",&n,&K);
	for(int i=1;i<=n;i++){
		scanf("%lld%d",&a[i],&b[i]);
		s[i]=s[i-1]+a[i];
		if(b[i]==1&&a[i]+a[i]>K){
			puts("-1");
			return 0;
		}
	}
	for(int i=1;i<=n;i++){
		if(b[i]==1){
			L[i]=((-2*s[i-1])%K+K)%K;
			R[i]=((-2*s[i])%K+K)%K;
		}else L[i]=0,R[i]=K-1;
	}
	for(int i=1;i<=n;i++) v[++m]=L[i],v[++m]=R[i];
	sort(v+1,v+m+1);
	m=unique(v+1,v+m+1)-v-1;
	for(int i=n;i;i--){
		L[i]=lower_bound(v+1,v+m+1,L[i])-v;
		R[i]=lower_bound(v+1,v+m+1,R[i])-v;
		f[i]=calc(L[i]);
		if(L[i]>R[i]) update(1,1,m,R[i]+1,L[i]-1,i);
		else update(1,1,m,1,L[i]-1,i),update(1,1,m,R[i]+1,m,i);
	}
	ll ans=1e18;
	for(int i=1;i<=m;i++){
		ans=min(ans,calc(i));
	}
	printf("%lld",ans+s[n]+s[n]);
	return 0;
}